Let a,b∈ℚabℚa,b\in\mathbb{Q} with φ⁢(m)=a+b⁢nφmabn\varphi(\sqrt{m})=a+b\sqrt{n}. Since φ⁢(a)=aφaa\varphi(a)=a and φφ\varphi is injective, b≠0b0b\neq 0. Also, m=φ⁢(m)=φ⁢((m)2)=(φ⁢(m))2=(a+b⁢n)2=a2+2⁢a⁢b⁢n+b2⁢nmφmφsuperscriptm2superscriptφm2superscriptabn2superscripta22abnsuperscriptb2nm=\varphi(m)=\varphi((\sqrt{m})^{2})=(\varphi(\sqrt{m}))^{2}=(a+b\sqrt{n})^{2}%
=a^{2}+2ab\sqrt{n}+b^{2}n. If a≠0a0a\neq 0, then n=m-a2-b2⁢n2⁢a⁢b∈ℚnmsuperscripta2superscriptb2n2abℚ\displaystyle\sqrt{n}=\frac{m-a^{2}-b^{2}n}{2ab}\in\mathbb{Q}, a contradiction. Thus, a=0a0a=0. Therefore, m=b2⁢nmsuperscriptb2nm=b^{2}n. Since mmm is squarefree, b2=1superscriptb21b^{2}=1. Hence, m=nmnm=n, a contradiction. It follows that KKK and LLL are not isomorphic.
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Corollary.

There are infinitely many distinct quadratic fields.

Proof.

Note that there are infinitely many elements of SSS. Moreover, if mmm and nnn are distinct elements of SSS, then ℚ⁢(m)ℚm\mathbb{Q}(\sqrt{m}) and ℚ⁢(n)ℚn\mathbb{Q}(\sqrt{n}) are not isomorphic and thus cannot be equal.
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